Unit 3 extends your differentiation skills to more complex functions. You'll learn how to:
A composite function is a function within a function: f(g(x)). The chain rule tells us how to differentiate these:
"Derivative of outside times derivative of inside"
Think of it as: "What's the derivative of the outside function, evaluated at the inside function, times the derivative of the inside function."
Find the derivative of f(x) = (x² + 3x)⁵
Solution:
Let g(x) = x² + 3x (inside function)
Let f(u) = u⁵ (outside function)
g'(x) = 2x + 3
f'(u) = 5u⁴
Using chain rule: f'(x) = f'(g(x)) · g'(x) = 5(x² + 3x)⁴ · (2x + 3)
Sometimes you need to apply the chain rule multiple times:
Find the derivative of f(x) = sin(cos(x²))
Solution:
Let u = x², v = cos(u), f(x) = sin(v)
du/dx = 2x
dv/du = -sin(u) = -sin(x²)
df/dv = cos(v) = cos(cos(x²))
Using chain rule: f'(x) = (df/dv)(dv/du)(du/dx)
= cos(cos(x²)) · (-sin(x²)) · 2x
= -2x sin(x²) cos(cos(x²))
When a function is defined implicitly (not solved for y), we use implicit differentiation:
Remember: d/dx[y] = dy/dx
When you see y in an equation, treat it as a function of x and use the chain rule: d/dx[y²] = 2y · dy/dx
Find dy/dx for x² + y² = 25
Solution:
Differentiate both sides with respect to x:
d/dx[x²] + d/dx[y²] = d/dx[25]
2x + 2y · dy/dx = 0
2y · dy/dx = -2x
dy/dx = -x/y
Find dy/dx for x³ + y³ = 3xy
Solution:
Differentiate both sides:
3x² + 3y² · dy/dx = 3y + 3x · dy/dx
3y² · dy/dx - 3x · dy/dx = 3y - 3x²
dy/dx(3y² - 3x) = 3y - 3x²
dy/dx = (3y - 3x²)/(3y² - 3x) = (y - x²)/(y² - x)
If f and g are inverse functions, then:
The derivative of an inverse function
To find the derivative of an inverse function, take the reciprocal of the derivative of the original function, evaluated at the inverse function.
If f(x) = x³ + 2x + 1, find (f⁻¹)'(4)
Solution:
First, find f'(x) = 3x² + 2
We need to find f⁻¹(4), which means f(x) = 4
x³ + 2x + 1 = 4 → x³ + 2x - 3 = 0 → x = 1
So f⁻¹(4) = 1
(f⁻¹)'(4) = 1/f'(f⁻¹(4)) = 1/f'(1) = 1/(3(1)² + 2) = 1/5
| Function | Inverse | Derivative of Inverse |
|---|---|---|
| sin(x) | arcsin(x) | 1/√(1-x²) |
| cos(x) | arccos(x) | -1/√(1-x²) |
| tan(x) | arctan(x) | 1/(1+x²) |
| eˣ | ln(x) | 1/x |
Use logarithmic differentiation when you have functions of the form f(x)^(g(x)) or complex products/quotients:
Find the derivative of f(x) = x^x
Solution:
Let y = x^x
Take natural log: ln(y) = ln(x^x) = x ln(x)
Differentiate implicitly: (1/y) · dy/dx = ln(x) + x · (1/x) = ln(x) + 1
Solve for dy/dx: dy/dx = y(ln(x) + 1) = x^x(ln(x) + 1)
Higher-order derivatives are derivatives of derivatives:
Second derivative
Third derivative
nth derivative
Find f''(x) for f(x) = x⁴ - 3x² + 2x
Solution:
f'(x) = 4x³ - 6x + 2
f''(x) = 12x² - 6
f'''(x) = 24x
f⁽⁴⁾(x) = 24
Find the derivative of f(x) = (x² + 1)³(2x - 3)⁴
Solution:
Use product rule with chain rule:
Let u = (x² + 1)³ and v = (2x - 3)⁴
u' = 3(x² + 1)² · 2x = 6x(x² + 1)²
v' = 4(2x - 3)³ · 2 = 8(2x - 3)³
f'(x) = u'v + uv' = 6x(x² + 1)²(2x - 3)⁴ + (x² + 1)³ · 8(2x - 3)³
= 2(x² + 1)²(2x - 3)³[3x(2x - 3) + 4(x² + 1)]
Find dy/dx for x²y + y³ = 6x
Solution:
Differentiate implicitly:
d/dx[x²y] + d/dx[y³] = d/dx[6x]
2xy + x² · dy/dx + 3y² · dy/dx = 6
dy/dx(x² + 3y²) = 6 - 2xy
dy/dx = (6 - 2xy)/(x² + 3y²)
Find the derivative of f(x) = arcsin(2x)
Solution:
Using the derivative of inverse sine with chain rule:
f'(x) = 1/√(1 - (2x)²) · d/dx[2x]
= 1/√(1 - 4x²) · 2
= 2/√(1 - 4x²)
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In Unit 4, you'll learn about applications of derivatives including optimization and related rates.
Go to Unit 4: Applications of Derivatives →